expectation of women sitting next to at least one man we are given that there are $5 \ men$ and $5 \ women$ are seated randomly in a single circle of chairs. What is the expected number of women sitting next to at least one man?
My approach
As we are given 10 seats, we can number them as $S_i$ where $ i∈(1,...,10)$
Here $S_i$ takes the value 1 if there is a woman sitting on seat number $i$ otherwise it takes the value 0.
we can define expectation as  $$E(X)= E(S_1)+...+E(S_{10})$$
to compute the required probability i have used the method
$$E(S_1)= \Pr(S_1 = 1)= 1-\Pr(S_1 = 0)$$ because by symmetry $E(S_1)=...=E(S_{10})$
answer is coming out to be $25/6$
Can I solve it in any other way?
 A: 1st Solution. To make amend of your approach, let $W_i$ be the event that a woman seats on position $i$, and then define $S_i$ as the indicator function
$$ S_i
= \mathbf{1}_{W_i \cap (W_{i-1} \cap W_{i+1})^c}
= \begin{cases}
1, & \begin{aligned} & \text{if a woman seats on position $i$} \\ & \text{and at least one man sits next to it,} \end{aligned} \\[0.5em]
0, & \text{otherwise.}
\end{cases} $$
Then the number $X$ of women sitting next to at least one man can be written by $X=\sum_{i=1}^{10} S_i$. Moreover, by the multiplication rule,
\begin{align*}
\mathbf{P}(S_i=1)
&= \mathbf{P}(W_i) \mathbf{P}((W_{i-1} \cap W_{i+1})^c \mid W_i) \\
&= \mathbf{P}(W_i) \bigl[ 1 -  \mathbf{P}(W_{i-1} \cap W_{i+1} \mid W_i) \bigr] \\
&= \mathbf{P}(W_i) \bigl[ 1 -  \mathbf{P}(W_{i-1} \mid W_i)\mathbf{P}(W_{i+1} \mid W_i \cap W_{i-1}) \bigr] \\
&= \tfrac{5}{10}\left(1-\tfrac{4}{9}\cdot\tfrac{3}{8}\right) \\
&= \tfrac{5}{12}.
\end{align*}
Therefore it follows that
$$ \mathbf{E}[X] = \sum_{i=1}^{10} \mathbf{P}(S_i=1) = \frac{25}{6}. $$
2nd Solution. Instead, suppose the women are numbered $1, 2, \ldots, 5$ and then let
$$ S'_i
= \begin{cases}
1, & \text{if woman $i$ seats next to at least one man,} \\
0, & \text{otherwise.}
\end{cases} $$
Then $X = \sum_{i=1}^{5} S'_i$, and
$$ \mathbf{P}(S'_i = 1) = 1 - \mathbf{P}(S'_i = 0) = 1 - \tfrac{4}{9}\cdot\tfrac{3}{8} = \tfrac{5}{6}.  $$
So the answer is
$$\mathbf{E}[X] = \sum_{i=1}^{5} \mathbf{P}(S'_i=1) = \frac{25}{6}. $$
